Recall the related notions of Lie groupoid, Lie algebroid, generalized morphism of Lie groupoids, and cohomology of Lie algebroid. Henceforth, I will drop the word "Lie" for all those things listed above, because I want to reuse it: there is a functor "Lie" from the 1-category of groupoids to the 1-category of algebroids. There is also a (contravariant) functor "cohomology" from the 1-category of algebroids to the 1-category of graded commutative algebras. However, their composition does not extend to generalized morphisms (which are really 1-morphisms in a 2-category of groupoids).

In particular, the algebroids for equivalent groupoids need not have isomorphic cohomology. A good example is as follows. For any manifold $M$, there is a "pair" groupoid $M\times M \rightrightarrows M$ with object the points in $M$ and a unique morphism between each pair of points. In fact, this groupoid is equivalent to the groupoid $\{\text{pt}\}$ with one object and one morphisms. But $\operatorname{Lie}(M\times M \rightrightarrows M) = {\rm T}M$ is the tangent groupoid, and the cohomology of this algebroid is the de Rham cohomology of $M$, which need not be trivial.

My question is:

If $G_1,G_2$ are two equivalent groupoids, and if both are source-simply-connected, does it follow that the cohomologies of the algebroids $\operatorname{Lie}(G_1)$ and $\operatorname{Lie}(G_2)$ are isomorphic?

There is a converse question, for which I am less optimistic, and that I haven't thought much about myself:

If $G_1,G_2$ are source-simply-connected groupoids and the cohomologies of $\operatorname{Lie}(G_1)$, $\operatorname{Lie}(G_2)$ are isomorphic, does it follow that $G_1,G_2$ are equivalent?

My motivation is the following. There ought to be (but there is not, although almost) a "Lie III theorem" that says that the categories of algebroids and of source-simply-connected groupoids are equivalent. Groupoids present stacks, and the question becomes what "stack-like" thing algebroids present. If the answers to both questions are "yes", then the "stack-like thing" presented by an algebroid just *is* its cohomology. But probably the answers are not both "yes" — even answers "yes, no" means that, well, the cohomology doesn't entirely determine the stack, but it is an invariant.